3.282 \(\int \frac{x^{11} \sqrt{c+d x^3}}{8 c-d x^3} \, dx\)

Optimal. Leaf size=111 \[ -\frac{1024 c^3 \sqrt{c+d x^3}}{3 d^4}-\frac{38 c^2 \left (c+d x^3\right )^{3/2}}{3 d^4}+\frac{1024 c^{7/2} \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )}{d^4}-\frac{4 c \left (c+d x^3\right )^{5/2}}{5 d^4}-\frac{2 \left (c+d x^3\right )^{7/2}}{21 d^4} \]

[Out]

(-1024*c^3*Sqrt[c + d*x^3])/(3*d^4) - (38*c^2*(c + d*x^3)^(3/2))/(3*d^4) - (4*c*(c + d*x^3)^(5/2))/(5*d^4) - (
2*(c + d*x^3)^(7/2))/(21*d^4) + (1024*c^(7/2)*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/d^4

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Rubi [A]  time = 0.0967581, antiderivative size = 111, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {446, 88, 50, 63, 206} \[ -\frac{1024 c^3 \sqrt{c+d x^3}}{3 d^4}-\frac{38 c^2 \left (c+d x^3\right )^{3/2}}{3 d^4}+\frac{1024 c^{7/2} \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )}{d^4}-\frac{4 c \left (c+d x^3\right )^{5/2}}{5 d^4}-\frac{2 \left (c+d x^3\right )^{7/2}}{21 d^4} \]

Antiderivative was successfully verified.

[In]

Int[(x^11*Sqrt[c + d*x^3])/(8*c - d*x^3),x]

[Out]

(-1024*c^3*Sqrt[c + d*x^3])/(3*d^4) - (38*c^2*(c + d*x^3)^(3/2))/(3*d^4) - (4*c*(c + d*x^3)^(5/2))/(5*d^4) - (
2*(c + d*x^3)^(7/2))/(21*d^4) + (1024*c^(7/2)*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/d^4

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{x^{11} \sqrt{c+d x^3}}{8 c-d x^3} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{x^3 \sqrt{c+d x}}{8 c-d x} \, dx,x,x^3\right )\\ &=\frac{1}{3} \operatorname{Subst}\left (\int \left (-\frac{57 c^2 \sqrt{c+d x}}{d^3}+\frac{512 c^3 \sqrt{c+d x}}{d^3 (8 c-d x)}-\frac{6 c (c+d x)^{3/2}}{d^3}-\frac{(c+d x)^{5/2}}{d^3}\right ) \, dx,x,x^3\right )\\ &=-\frac{38 c^2 \left (c+d x^3\right )^{3/2}}{3 d^4}-\frac{4 c \left (c+d x^3\right )^{5/2}}{5 d^4}-\frac{2 \left (c+d x^3\right )^{7/2}}{21 d^4}+\frac{\left (512 c^3\right ) \operatorname{Subst}\left (\int \frac{\sqrt{c+d x}}{8 c-d x} \, dx,x,x^3\right )}{3 d^3}\\ &=-\frac{1024 c^3 \sqrt{c+d x^3}}{3 d^4}-\frac{38 c^2 \left (c+d x^3\right )^{3/2}}{3 d^4}-\frac{4 c \left (c+d x^3\right )^{5/2}}{5 d^4}-\frac{2 \left (c+d x^3\right )^{7/2}}{21 d^4}+\frac{\left (1536 c^4\right ) \operatorname{Subst}\left (\int \frac{1}{(8 c-d x) \sqrt{c+d x}} \, dx,x,x^3\right )}{d^3}\\ &=-\frac{1024 c^3 \sqrt{c+d x^3}}{3 d^4}-\frac{38 c^2 \left (c+d x^3\right )^{3/2}}{3 d^4}-\frac{4 c \left (c+d x^3\right )^{5/2}}{5 d^4}-\frac{2 \left (c+d x^3\right )^{7/2}}{21 d^4}+\frac{\left (3072 c^4\right ) \operatorname{Subst}\left (\int \frac{1}{9 c-x^2} \, dx,x,\sqrt{c+d x^3}\right )}{d^4}\\ &=-\frac{1024 c^3 \sqrt{c+d x^3}}{3 d^4}-\frac{38 c^2 \left (c+d x^3\right )^{3/2}}{3 d^4}-\frac{4 c \left (c+d x^3\right )^{5/2}}{5 d^4}-\frac{2 \left (c+d x^3\right )^{7/2}}{21 d^4}+\frac{1024 c^{7/2} \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )}{d^4}\\ \end{align*}

Mathematica [A]  time = 0.0749603, size = 81, normalized size = 0.73 \[ \frac{107520 c^{7/2} \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )-2 \sqrt{c+d x^3} \left (764 c^2 d x^3+18632 c^3+57 c d^2 x^6+5 d^3 x^9\right )}{105 d^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^11*Sqrt[c + d*x^3])/(8*c - d*x^3),x]

[Out]

(-2*Sqrt[c + d*x^3]*(18632*c^3 + 764*c^2*d*x^3 + 57*c*d^2*x^6 + 5*d^3*x^9) + 107520*c^(7/2)*ArcTanh[Sqrt[c + d
*x^3]/(3*Sqrt[c])])/(105*d^4)

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Maple [C]  time = 0.056, size = 582, normalized size = 5.2 \begin{align*} -{\frac{1}{d} \left ({\frac{2\,{x}^{9}}{21}\sqrt{d{x}^{3}+c}}+{\frac{2\,c{x}^{6}}{105\,d}\sqrt{d{x}^{3}+c}}-{\frac{8\,{c}^{2}{x}^{3}}{315\,{d}^{2}}\sqrt{d{x}^{3}+c}}+{\frac{16\,{c}^{3}}{315\,{d}^{3}}\sqrt{d{x}^{3}+c}} \right ) }-8\,{\frac{c}{{d}^{2}} \left ( 2/15\,{x}^{6}\sqrt{d{x}^{3}+c}+{\frac{2\,c{x}^{3}\sqrt{d{x}^{3}+c}}{45\,d}}-{\frac{4\,{c}^{2}\sqrt{d{x}^{3}+c}}{45\,{d}^{2}}} \right ) }-{\frac{128\,{c}^{2}}{9\,{d}^{4}} \left ( d{x}^{3}+c \right ) ^{{\frac{3}{2}}}}-512\,{\frac{{c}^{3}}{{d}^{3}} \left ( 2/3\,{\frac{\sqrt{d{x}^{3}+c}}{d}}+{\frac{i/3\sqrt{2}}{{d}^{3}}\sum _{{\it \_alpha}={\it RootOf} \left ( d{{\it \_Z}}^{3}-8\,c \right ) }{\frac{\sqrt [3]{-{d}^{2}c} \left ( i\sqrt [3]{-{d}^{2}c}{\it \_alpha}\,\sqrt{3}d-i\sqrt{3} \left ( -{d}^{2}c \right ) ^{2/3}+2\,{{\it \_alpha}}^{2}{d}^{2}-\sqrt [3]{-{d}^{2}c}{\it \_alpha}\,d- \left ( -{d}^{2}c \right ) ^{2/3} \right ) }{\sqrt{d{x}^{3}+c}}\sqrt{{\frac{i/2d}{\sqrt [3]{-{d}^{2}c}} \left ( 2\,x+{\frac{-i\sqrt{3}\sqrt [3]{-{d}^{2}c}+\sqrt [3]{-{d}^{2}c}}{d}} \right ) }}\sqrt{{\frac{d}{-3\,\sqrt [3]{-{d}^{2}c}+i\sqrt{3}\sqrt [3]{-{d}^{2}c}} \left ( x-{\frac{\sqrt [3]{-{d}^{2}c}}{d}} \right ) }}\sqrt{{\frac{-i/2d}{\sqrt [3]{-{d}^{2}c}} \left ( 2\,x+{\frac{i\sqrt{3}\sqrt [3]{-{d}^{2}c}+\sqrt [3]{-{d}^{2}c}}{d}} \right ) }}{\it EllipticPi} \left ( 1/3\,\sqrt{3}\sqrt{{\frac{id\sqrt{3}}{\sqrt [3]{-{d}^{2}c}} \left ( x+1/2\,{\frac{\sqrt [3]{-{d}^{2}c}}{d}}-{\frac{i/2\sqrt{3}\sqrt [3]{-{d}^{2}c}}{d}} \right ) }},-1/18\,{\frac{2\,i\sqrt [3]{-{d}^{2}c}\sqrt{3}{{\it \_alpha}}^{2}d-i \left ( -{d}^{2}c \right ) ^{2/3}\sqrt{3}{\it \_alpha}+i\sqrt{3}cd-3\, \left ( -{d}^{2}c \right ) ^{2/3}{\it \_alpha}-3\,cd}{cd}},\sqrt{{\frac{i\sqrt{3}\sqrt [3]{-{d}^{2}c}}{d} \left ( -3/2\,{\frac{\sqrt [3]{-{d}^{2}c}}{d}}+{\frac{i/2\sqrt{3}\sqrt [3]{-{d}^{2}c}}{d}} \right ) ^{-1}}} \right ) }} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^11*(d*x^3+c)^(1/2)/(-d*x^3+8*c),x)

[Out]

-1/d*(2/21*x^9*(d*x^3+c)^(1/2)+2/105/d*c*x^6*(d*x^3+c)^(1/2)-8/315*c^2/d^2*x^3*(d*x^3+c)^(1/2)+16/315*c^3*(d*x
^3+c)^(1/2)/d^3)-8*c/d^2*(2/15*x^6*(d*x^3+c)^(1/2)+2/45/d*c*x^3*(d*x^3+c)^(1/2)-4/45*c^2*(d*x^3+c)^(1/2)/d^2)-
128/9*c^2*(d*x^3+c)^(3/2)/d^4-512*c^3/d^3*(2/3*(d*x^3+c)^(1/2)/d+1/3*I/d^3*2^(1/2)*sum((-d^2*c)^(1/3)*(1/2*I*d
*(2*x+1/d*(-I*3^(1/2)*(-d^2*c)^(1/3)+(-d^2*c)^(1/3)))/(-d^2*c)^(1/3))^(1/2)*(d*(x-1/d*(-d^2*c)^(1/3))/(-3*(-d^
2*c)^(1/3)+I*3^(1/2)*(-d^2*c)^(1/3)))^(1/2)*(-1/2*I*d*(2*x+1/d*(I*3^(1/2)*(-d^2*c)^(1/3)+(-d^2*c)^(1/3)))/(-d^
2*c)^(1/3))^(1/2)/(d*x^3+c)^(1/2)*(I*(-d^2*c)^(1/3)*_alpha*3^(1/2)*d-I*3^(1/2)*(-d^2*c)^(2/3)+2*_alpha^2*d^2-(
-d^2*c)^(1/3)*_alpha*d-(-d^2*c)^(2/3))*EllipticPi(1/3*3^(1/2)*(I*(x+1/2/d*(-d^2*c)^(1/3)-1/2*I*3^(1/2)/d*(-d^2
*c)^(1/3))*3^(1/2)*d/(-d^2*c)^(1/3))^(1/2),-1/18/d*(2*I*(-d^2*c)^(1/3)*3^(1/2)*_alpha^2*d-I*(-d^2*c)^(2/3)*3^(
1/2)*_alpha+I*3^(1/2)*c*d-3*(-d^2*c)^(2/3)*_alpha-3*c*d)/c,(I*3^(1/2)/d*(-d^2*c)^(1/3)/(-3/2/d*(-d^2*c)^(1/3)+
1/2*I*3^(1/2)/d*(-d^2*c)^(1/3)))^(1/2)),_alpha=RootOf(_Z^3*d-8*c)))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11*(d*x^3+c)^(1/2)/(-d*x^3+8*c),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.80048, size = 420, normalized size = 3.78 \begin{align*} \left [\frac{2 \,{\left (26880 \, c^{\frac{7}{2}} \log \left (\frac{d x^{3} + 6 \, \sqrt{d x^{3} + c} \sqrt{c} + 10 \, c}{d x^{3} - 8 \, c}\right ) -{\left (5 \, d^{3} x^{9} + 57 \, c d^{2} x^{6} + 764 \, c^{2} d x^{3} + 18632 \, c^{3}\right )} \sqrt{d x^{3} + c}\right )}}{105 \, d^{4}}, -\frac{2 \,{\left (53760 \, \sqrt{-c} c^{3} \arctan \left (\frac{\sqrt{d x^{3} + c} \sqrt{-c}}{3 \, c}\right ) +{\left (5 \, d^{3} x^{9} + 57 \, c d^{2} x^{6} + 764 \, c^{2} d x^{3} + 18632 \, c^{3}\right )} \sqrt{d x^{3} + c}\right )}}{105 \, d^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11*(d*x^3+c)^(1/2)/(-d*x^3+8*c),x, algorithm="fricas")

[Out]

[2/105*(26880*c^(7/2)*log((d*x^3 + 6*sqrt(d*x^3 + c)*sqrt(c) + 10*c)/(d*x^3 - 8*c)) - (5*d^3*x^9 + 57*c*d^2*x^
6 + 764*c^2*d*x^3 + 18632*c^3)*sqrt(d*x^3 + c))/d^4, -2/105*(53760*sqrt(-c)*c^3*arctan(1/3*sqrt(d*x^3 + c)*sqr
t(-c)/c) + (5*d^3*x^9 + 57*c*d^2*x^6 + 764*c^2*d*x^3 + 18632*c^3)*sqrt(d*x^3 + c))/d^4]

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Sympy [A]  time = 155.865, size = 99, normalized size = 0.89 \begin{align*} \frac{2 \left (- \frac{512 c^{4} \operatorname{atan}{\left (\frac{\sqrt{c + d x^{3}}}{3 \sqrt{- c}} \right )}}{\sqrt{- c}} - \frac{512 c^{3} \sqrt{c + d x^{3}}}{3} - \frac{19 c^{2} \left (c + d x^{3}\right )^{\frac{3}{2}}}{3} - \frac{2 c \left (c + d x^{3}\right )^{\frac{5}{2}}}{5} - \frac{\left (c + d x^{3}\right )^{\frac{7}{2}}}{21}\right )}{d^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**11*(d*x**3+c)**(1/2)/(-d*x**3+8*c),x)

[Out]

2*(-512*c**4*atan(sqrt(c + d*x**3)/(3*sqrt(-c)))/sqrt(-c) - 512*c**3*sqrt(c + d*x**3)/3 - 19*c**2*(c + d*x**3)
**(3/2)/3 - 2*c*(c + d*x**3)**(5/2)/5 - (c + d*x**3)**(7/2)/21)/d**4

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Giac [A]  time = 1.09088, size = 135, normalized size = 1.22 \begin{align*} -\frac{1024 \, c^{4} \arctan \left (\frac{\sqrt{d x^{3} + c}}{3 \, \sqrt{-c}}\right )}{\sqrt{-c} d^{4}} - \frac{2 \,{\left (5 \,{\left (d x^{3} + c\right )}^{\frac{7}{2}} d^{24} + 42 \,{\left (d x^{3} + c\right )}^{\frac{5}{2}} c d^{24} + 665 \,{\left (d x^{3} + c\right )}^{\frac{3}{2}} c^{2} d^{24} + 17920 \, \sqrt{d x^{3} + c} c^{3} d^{24}\right )}}{105 \, d^{28}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11*(d*x^3+c)^(1/2)/(-d*x^3+8*c),x, algorithm="giac")

[Out]

-1024*c^4*arctan(1/3*sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*d^4) - 2/105*(5*(d*x^3 + c)^(7/2)*d^24 + 42*(d*x^3 +
c)^(5/2)*c*d^24 + 665*(d*x^3 + c)^(3/2)*c^2*d^24 + 17920*sqrt(d*x^3 + c)*c^3*d^24)/d^28